Question

An important application of regression analysis in accounting is in the estimation of cost. By collecting data on volume and cost and using the least squares method to develop an estimated regression equation relating volume and cost, an accountant can estimate the cost associated with a particular manufacturing volume. Consider the following sample of production volumes and total cost data for a manufacturing operation. Production Volume (units) Total Cost ($) 400 4,600 450 5,600 550 6,000 600 6,500 700 7,000 750 7,600 Compute b1 and b0 (to 1 decimal). b1   b0   Complete the estimated regression equation (to 1 decimal). =  +  x What is the variable cost per unit produced (to 1 decimal)? $ Compute the coefficient of determination (to 3 decimals). Note: report r2 between 0 and 1. r2 =   What percentage of the variation in total cost can be explained by the production volume (to 1 decimal)? % The company's production schedule shows 500 units must be produced next month. What is the estimated total cost for this operation (to the nearest whole number)?

Answer 1

a.

Line of Regression Y on X i.e Y = bo + b1 X
X	Y	(Xi - Mean)^2	(Yi - Mean)^2	(Xi-Mean)*(Yi-Mean)
400	4600	30625	2613611.2	282916.7
450	5600	15625	380277.8	77083.3
550	6000	625	46944.5	5416.7
600	6500	625	80277.8	7083.3
700	7000	15625	613611.1	97916.7
750	7600	30625	1913611	242083.3

calculation procedure for regression
mean of X = ∑ X / n = 575
mean of Y = ∑ Y / n = 6216.6667
∑ (Xi - Mean)^2 = 93750
∑ (Yi - Mean)^2 = 5648333.4
∑ (Xi-Mean)*(Yi-Mean) = 712500
b1 = ∑ (Xi-Mean)*(Yi-Mean) / ∑ (Xi - Mean)^2
= 712500 / 93750
= 7.6
bo = ∑ Y / n - b1 * ∑ X / n
bo = 6216.6667 - 7.6*575 = 1846.7
value of regression equation is, Y = bo + b1 X
Y'=1846.7+7.6* X          

b.

Y'=1846.7+7.6* X
the variable cost per unit produced = 7.6$

c.

( X)	( Y)	X^2	Y^2	X*Y
400	4600	160000	2.1E+07	1840000
450	5600	202500	3.1E+07	2520000
550	6000	302500	3.6E+07	3300000
600	6500	360000	4.2E+07	3900000
700	7000	490000	4.9E+07	4900000
750	7600	562500	5.8E+07	5700000

calculation procedure for correlation
sum of (x) = ∑x = 3450
sum of (y) = ∑y = 37300
sum of (x^2)= ∑x^2 = 2077500
sum of (y^2)= ∑y^2 = 237530000
sum of (x*y)= ∑x*y = 22160000
to caluclate value of r( x,y) = covariance ( x,y ) / sd (x) * sd (y)
covariance ( x,y ) = [ ∑x*y - N *(∑x/N) * (∑y/N) ]/n-1
= 22160000 - [ 6 * (3450/6) * (37300/6) ]/6- 1
= 118750
and now to calculate r( x,y) = 118750/ (SQRT(1/6*22160000-(1/6*3450)^2) ) * ( SQRT(1/6*22160000-(1/6*37300)^2)
=118750 / (125*970.3)
=1
value of correlation is =1
coefficient of determination = r^2 = 1
properties of correlation
1. If r = 1 Corrlation is called Perfect Positive Correlation
2. If r = -1 Correlation is called Perfect Negative Correlation
3. If r = 0 Correlation is called Zero Correlation
& with above we conclude that correlation ( r ) is = 0.9791> 0 ,perfect positive correlation              
percentage of the variation in total cost can be explained by the production volume = 97.91%

d.

The company's production schedule shows 500 units must be produced next month.
Y'=1846.7+7.6* X
Y'=1846.7+(7.6*500) = 5646.7
estimated total cost for this operation = 5647